Two capillary of length L and 2L and of radius R and 2R are connected in series. The net rate of flow of fluid through them will be (given rate of the flow through single capillary, $\mathrm{X}={\mathrm{\pi PR}}^4/8\mathrm{\eta L})$

Fluid resistance is given by $\mathrm{R}=\frac{8\mathrm{\eta l}}{{\mathrm{\pi r}}^4}.$
When two capillary tubes of same size are joined in parallel, then equivalent fluid resistance is 
 ${\mathrm{R}}_{\mathrm{e}}={\mathrm{R}}_1+{\mathrm{R}}_2=\frac{8\mathrm{\eta L}}{{\mathrm{\pi r}}^4}+\frac{8\mathrm{\eta }\times 2\mathrm{L}}{\mathrm{\pi }{\left(2\mathrm{R}\right)}^4}=\left(\frac{8\mathrm{\eta L}}{{\mathrm{\pi r}}^4}\right)\times \frac{9}{8}$
Equivalent resistance becomes $\frac{9}{8}$ times so rate of flow will be  $\frac{8}{9}\mathrm{X}$